An arbitrarily shaped rigid rotor is a
rigid body of arbitrary shape with its
center of mass fixed (or in uniform rectilinear motion) in field-free space
R3, so that its energy consists only of rotational kinetic energy (and possibly constant translational energy that can be ignored). A rigid body can be (partially) characterized by the three eigenvalues of its
moment of inertia tensor, which are real nonnegative values known as
principal moments of inertia. In
microwave spectroscopy—the spectroscopy based on rotational transitions—one usually classifies molecules (seen as rigid rotors) as follows: • spherical rotors • symmetric rotors • oblate symmetric rotors • prolate symmetric rotors • asymmetric rotors This classification depends on the
relative magnitudes of the principal moments of inertia.
Coordinates of the rigid rotor Different branches of physics and engineering use different coordinates for the description of the kinematics of a rigid rotor. In molecular physics
Euler angles are used almost exclusively. In quantum mechanical applications it is advantageous to use Euler angles in a convention that is a
simple extension of the physical convention of
spherical polar coordinates. The first step is the attachment of a
right-handed orthonormal frame (3-dimensional system of orthogonal axes) to the rotor (a
body-fixed frame). This frame can be attached arbitrarily to the body, but often one uses the principal axes frame—the normalized eigenvectors of the inertia
tensor, which always can be chosen orthonormal, since the tensor is
symmetric. When the rotor possesses a symmetry-axis, it usually coincides with one of the principal axes. It is convenient to choose as body-fixed
z-axis the highest-order symmetry axis. One starts by aligning the body-fixed frame with a
space-fixed frame (laboratory axes), so that the body-fixed
x,
y, and
z axes coincide with the space-fixed
X,
Y, and
Z axis. Secondly, the body and its frame are rotated
actively over a
positive angle \alpha\, around the
z-axis (by the
right-hand rule), which moves the y- to the y'-axis. Thirdly, one rotates the body and its frame over a positive angle \beta\, around the y'-axis. The
z-axis of the body-fixed frame has after these two rotations the longitudinal angle \alpha \, (commonly designated by \varphi\,) and the colatitude angle \beta\, (commonly designated by \theta\,), both with respect to the space-fixed frame. If the rotor were cylindrical symmetric around its
z-axis, like the linear rigid rotor, its orientation in space would be unambiguously specified at this point. If the body lacks cylinder (axial) symmetry, a last rotation around its
z-axis (which has polar coordinates \beta\, and \alpha\,) is necessary to specify its orientation completely. Traditionally the last rotation angle is called \gamma\,. The
convention for Euler angles described here is known as the z''-y'-z convention; it can be shown (in the same manner as in
this article) that it is equivalent to the z-y-z convention in which the order of rotations is reversed. The total matrix of the three consecutive rotations is the product \mathbf{R}(\alpha,\beta,\gamma)= \begin{pmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \\ \end{pmatrix} \begin{pmatrix} \cos\gamma & -\sin\gamma & 0 \\ \sin\gamma & \cos\gamma & 0 \\ 0 & 0 & 1 \end{pmatrix} Let \mathbf{r}(0) be the
coordinate vector of an arbitrary point \mathcal{P} in the body with respect to the body-fixed frame. The elements of \mathbf{r}(0) are the 'body-fixed coordinates' of \mathcal{P}. Initially \mathbf{r}(0) is also the space-fixed coordinate vector of \mathcal{P}. Upon rotation of the body, the body-fixed coordinates of \mathcal{P} do not change, but the space-fixed coordinate vector of \mathcal{P} becomes, \mathbf{r}(\alpha,\beta,\gamma)= \mathbf{R}(\alpha,\beta,\gamma)\mathbf{r}(0). In particular, if \mathcal{P} is initially on the space-fixed
Z-axis, it has the space-fixed coordinates \mathbf{R}(\alpha,\beta,\gamma) \begin{pmatrix} 0 \\ 0 \\ r \\ \end{pmatrix}= \begin{pmatrix} r \cos\alpha\sin\beta \\ r \sin\alpha \sin\beta \\ r \cos\beta \\ \end{pmatrix}, which shows the correspondence with the
spherical polar coordinates (in the physical convention). Knowledge of the Euler angles as function of time
t and the initial coordinates \mathbf{r}(0) determine the kinematics of the rigid rotor.
Classical Kinetic Energy The following text forms a generalization of the well-known special case of the
rotational energy of an object that rotates around
one axis. It will be assumed from here on that the body-fixed frame is a principal axes frame; it diagonalizes the instantaneous
inertia tensor \mathbf{I}(t) (expressed with respect to the space-fixed frame), i.e., \mathbf{R}(\alpha,\beta,\gamma)^{-1}\; \mathbf{I}(t)\; \mathbf{R}(\alpha,\beta,\gamma) = \mathbf{I}(0)\quad\hbox{with}\quad \mathbf{I}(0) = \begin{pmatrix} I_1 & 0 & 0 \\ 0 & I_2 & 0 \\ 0 & 0 & I_3 \\ \end{pmatrix}, where the Euler angles are time-dependent and in fact determine the time dependence of \mathbf{I}(t) by the inverse of this equation. This notation implies that at t=0 the Euler angles are zero, so that at t=0 the body-fixed frame coincides with the space-fixed frame. The classical kinetic energy
T of the rigid rotor can be expressed in different ways: • as a function of angular velocity • in Lagrangian form • as a function of angular momentum • in Hamiltonian form. Since each of these forms has its use and can be found in textbooks we will present all of them.
Angular velocity form As a function of angular velocity
T reads, T = \tfrac{1}{2} \left[ I_1 \omega_x^2 + I_2 \omega_y^2+ I_3 \omega_z^2 \right] with \begin{pmatrix} \omega_x \\ \omega_y \\ \omega_z \\ \end{pmatrix} = \begin{pmatrix} -\sin\beta\cos\gamma & \sin\gamma & 0 \\ \sin\beta\sin\gamma & \cos\gamma & 0 \\ \cos\beta & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} \dot{\alpha} \\ \dot{\beta} \\ \dot{\gamma} \\ \end{pmatrix}. The vector \boldsymbol{\omega} = (\omega_x, \omega_y, \omega_z) on the left hand side contains the components of the
angular velocity of the rotor expressed with respect to the body-fixed frame. The angular velocity satisfies equations of motion known as
Euler's equations (with zero applied torque, since by assumption the rotor is in field-free space). It can be shown that \boldsymbol{\omega} is
not the time derivative of any vector, in contrast to the usual
definition of velocity. The dots over the time-dependent Euler angles on the right hand side indicate
time derivatives. Note that a different
rotation matrix would result from a different choice of Euler angle convention used.
Lagrange form Back substitution of the expression of \boldsymbol{\omega} into
T gives the kinetic energy in
Lagrange form (as a function of the time derivatives of the Euler angles). In matrix-vector notation, 2 T = \begin{pmatrix} \dot{\alpha} & \dot{\beta} & \dot{\gamma} \end{pmatrix} \; \mathbf{g} \; \begin{pmatrix} \dot{\alpha} \\ \dot{\beta} \\ \dot{\gamma}\\ \end{pmatrix}, where \mathbf{g} is the metric tensor expressed in Euler angles—a non-orthogonal system of
curvilinear coordinates— \mathbf{g}= \begin{pmatrix} I_1 \sin^2\beta \cos^2\gamma+I_2\sin^2\beta\sin^2\gamma+I_3\cos^2\beta & (I_2-I_1) \sin\beta\sin\gamma\cos\gamma & I_3\cos\beta \\ (I_2-I_1) \sin\beta\sin\gamma\cos\gamma & I_1\sin^2\gamma+I_2\cos^2\gamma & 0 \\ I_3\cos\beta & 0 & I_3 \\ \end{pmatrix}.
Angular momentum form Often the kinetic energy is written as a function of the
angular momentum \mathbf{L} of the rigid rotor. With respect to the body-fixed frame it has the components L_i, and can be shown to be related to the angular velocity, \mathbf{L} = \mathbf{I}(0)\; \boldsymbol{\omega}\quad\hbox{or}\quad L_i = \frac{\partial T}{\partial\omega_i},\;\; i=x,\,y,\,z. This angular momentum is a conserved (time-independent) quantity if viewed from a stationary space-fixed frame. Since the body-fixed frame moves (depends on time) the components L_i are
not time independent. If we were to represent \mathbf{L} with respect to the stationary space-fixed frame, we would find time independent expressions for its components. The kinetic energy is expressed in terms of the angular momentum by T = \frac{1}{2} \left[ \frac{L_x^2}{I_1} + \frac{L_y^2}{I_2}+ \frac{L_z^2}{I_3}\right].
Hamilton form The
Hamilton form of the kinetic energy is written in terms of generalized momenta \begin{pmatrix} p_\alpha \\ p_\beta \\ p_\gamma \\ \end{pmatrix} \mathrel\stackrel{\mathrm{def}}{=} \begin{pmatrix} \partial T/{\partial \dot{\alpha}}\\ \partial T/{\partial \dot{\beta}} \\ \partial T/{\partial \dot{\gamma}} \\ \end{pmatrix} = \mathbf{g} \begin{pmatrix} \; \, \dot{\alpha} \\ \dot{\beta} \\ \dot{\gamma}\\ \end{pmatrix}, where it is used that the \mathbf{g} is symmetric. In Hamilton form the kinetic energy is, 2 T = \begin{pmatrix} p_{\alpha} & p_{\beta} & p_{\gamma} \end{pmatrix} \; \mathbf{g}^{-1} \; \begin{pmatrix} p_{\alpha} \\ p_{\beta} \\ p_{\gamma}\\ \end{pmatrix}, with the inverse metric tensor given by \sin^2\beta\; \mathbf{g}^{-1} = \begin{pmatrix} \frac{1}{I_1}\cos^2\gamma + \frac{1}{I_2}\sin^2\gamma & \left(\frac{1}{I_2} - \frac{1}{I_1}\right)\sin\beta\sin\gamma\cos\gamma & -\frac{1}{I_1}\cos\beta\cos^2\gamma - \frac{1}{I_2}\cos\beta\sin^2\gamma \\ \left(\frac{1}{I_2} - \frac{1}{I_1}\right)\sin\beta\sin\gamma\cos\gamma & \frac{1}{I_1}\sin^2\beta\sin^2\gamma + \frac{1}{I_2}\sin^2\beta\cos^2\gamma & \left(\frac{1}{I_1} - \frac{1}{I_2}\right)\sin\beta\cos\beta\sin\gamma\cos\gamma \\ -\frac{1}{I_1}\cos\beta\cos^2\gamma - \frac{1}{I_2}\cos\beta\sin^2\gamma & \left(\frac{1}{I_1} - \frac{1}{I_2}\right)\sin\beta\cos\beta\sin\gamma\cos\gamma & \frac{1}{I_1}\cos^2\beta\cos^2\gamma + \frac{1}{I_2}\cos^2\beta\sin^2\gamma + \frac{1}{I_3}\sin^2\beta \\ \end{pmatrix}. This inverse tensor is needed to obtain the
Laplace-Beltrami operator, which (multiplied by -\hbar^2) gives the quantum mechanical energy operator of the rigid rotor. The classical Hamiltonian given above can be rewritten to the following expression, which is needed in the phase integral arising in the classical
statistical mechanics of rigid rotors, \begin{align} T ={} &\frac{1}{2I_1 \sin^2\beta} \left( \left(p_\alpha - p_\gamma\cos\beta\right) \cos\gamma - p_\beta\sin\beta\sin\gamma \right)^2 +{} \\ &\frac{1}{2I_2 \sin^2\beta} \left( \left(p_\alpha - p_\gamma\cos\beta\right) \sin\gamma + p_\beta\sin\beta\cos\gamma \right)^2 + \frac{p_\gamma^2}{2I_3}. \\ \end{align}
Quantum mechanical rigid rotor As usual quantization is performed by the replacement of the generalized momenta by operators that give first derivatives with respect to its
canonically conjugate variables (positions). Thus, p_\alpha \longrightarrow -i \hbar \frac{\partial}{\partial \alpha} and similarly for p_\beta and p_\gamma. It is remarkable that this rule replaces the fairly complicated function p_\alpha of all three Euler angles, time derivatives of Euler angles, and inertia moments (characterizing the rigid rotor) by a simple
differential operator that does not depend on time or inertia moments and differentiates to one Euler angle only. The quantization rule is sufficient to obtain the operators that correspond with the classical angular momenta. There are two kinds: space-fixed and body-fixed angular momentum operators. Both are vector operators, i.e., both have three components that transform as vector components among themselves upon rotation of the space-fixed and the body-fixed frame, respectively. The explicit form of the rigid rotor angular momentum operators is given
here (but beware, they must be multiplied with \hbar). The body-fixed angular momentum operators are written as \hat{\mathcal{P}}_i. They satisfy
anomalous commutation relations. The quantization rule is
not sufficient to obtain the kinetic energy operator from the classical Hamiltonian. Since classically p_\beta commutes with \cos\beta and \sin\beta and the inverses of these functions, the position of these
trigonometric functions in the classical Hamiltonian is arbitrary. After quantization the commutation does no longer hold and the order of operators and functions in the Hamiltonian (energy operator) becomes a point of concern. Podolsky (for the special case of the symmetric rotor). This is one of the few cases where the Schrödinger equation can be solved analytically. All these cases were solved within a year of the formulation of the Schrödinger equation.) Nowadays it is common to proceed as follows. It can be shown that \hat{H} can be expressed in body-fixed angular momentum operators (in this proof one must carefully commute differential operators with trigonometric functions). The result has the same appearance as the classical formula expressed in body-fixed coordinates, \hat{H} = \frac{1}{2}\left[ \frac{\mathcal{P}_x^2}{I_1} + \frac{\mathcal{P}_y^2}{I_2} + \frac{\mathcal{P}_z^2}{I_3} \right]. The action of the \hat{\mathcal{P}}_i on the
Wigner D-matrix is simple. In particular \mathcal{P}^2\, D^j_{m'm}(\alpha,\beta,\gamma)^* = \hbar^2 j(j+1) D^j_{m'm}(\alpha,\beta,\gamma)^* \quad\hbox{with}\quad \mathcal{P}^2 = \mathcal{P}^2_x + \mathcal{P}_y^2+ \mathcal{P}_z^2, so that the Schrödinger equation for the spherical rotor (I=I_1=I_2=I_3) is solved with the (2j+1)^2 degenerate energy equal to \tfrac{\hbar^2 j(j+1)}{2I}. The symmetric top (= symmetric rotor) is characterized by I_1=I_2. It is a
prolate (cigar shaped) top if I_3 . In the latter case we write the Hamiltonian as \hat{H} = \frac{1}{2}\left[ \frac{\mathcal{P}^2}{I_1} + \mathcal{P}_z^2\left(\frac{1}{I_3} -\frac{1}{I_1} \right) \right], and use that \mathcal{P}_z^2\, D^j_{m k}(\alpha,\beta,\gamma)^* = \hbar^2 k^2\, D^j_{m k}(\alpha,\beta,\gamma)^*. Hence \hat{H}\,D^j_{m k}(\alpha,\beta,\gamma)^* = E_{jk} D^j_{m k}(\alpha,\beta,\gamma)^* with \frac{1}{\hbar^2}E_{jk} = \frac{j(j + 1)}{2I_1} + k^2\left(\frac{1}{2I_3} - \frac{1}{2I_1}\right). The eigenvalue E_{j0} is 2j+1-fold degenerate, for all eigenfunctions with m = -j, -j+1, \dots, j have the same eigenvalue. The energies with |k| > 0 are 2(2j+1)-fold degenerate. This exact solution of the Schrödinger equation of the symmetric top was first found in 1927. The asymmetric top problem ( I_1 \ne I_2 \ne I_3 ) is not soluble analytically. == Direct experimental observation of molecular rotations ==